So here's a question I had. Let's say I have a quantum mechanical system which obeys the Schrodinger equation.

$$ \hat H \psi = \hat T \psi + \hat V \psi $$

where $\hat H$ is the Hamiltonian, $\hat T \psi $ is the kinetic energy and $\hat V \psi $ is the potential. Let's say I wish to use this system as a quantum computer to perform some calculation. Now, let's I put the quantum computer itself into superposition? Now one may (reasonably) argue against this: How does one put the Hamiltonian this into superposition?

Quantum Field Theory! If I second quantize the Hamiltonian then I can have of quantum mechanical theory with superimposed particle numbers.


Now that I've superimposed the quantum computer. Is there a relation between the computation done by the original quantum computer and the super imposed quantum computer? Can someone refer me a paper where such an exploration has been done?

  • $\begingroup$ When you act on a state with a Hamiltonian for a given time, you take one state and transform it into another. Now does your question refer to 1) putting this final state into a superposition? That would just require operating on the state again to put it in superposition. Or is it 2) you put the physical quantum computer itself into a superposition. That is once again putting a state into superposition, except this time your state is every atom that makes up your physical quantum computer. So you would have to expand your wave function to include all those particles. $\endgroup$ Aug 15, 2023 at 7:57
  • $\begingroup$ I'm talking about 2 $\endgroup$ Aug 15, 2023 at 7:58
  • $\begingroup$ OK, so as mentioned, in order to put your physical quantum computer into a superposition, you would have to first define the state that includes all the particles that make up your physical computer (i.e. expand your Hilbert space). Then you can just treat it as its own quantum mechanical system (you don't need quantum field theory). Does that make sense? $\endgroup$ Aug 15, 2023 at 8:01
  • 2
    $\begingroup$ at the end of the day, a "quantum computer" is nothing but some quantum system, with the added implicit assumption that you are particularly good at controlling it. So the state of the quantum computer will always be "in a superposition" during any computation (as any quantum state is always writable as a superposition of other states in infinitely many ways) $\endgroup$
    – glS
    Aug 15, 2023 at 10:06
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    $\begingroup$ I'm identifying the quantum computer with the Hamiltonian rather than the ket. $\endgroup$ Aug 15, 2023 at 13:04

1 Answer 1


You're looking for Quantum computations without definite causal structure, where the order of the gates acting on your quantum system is controlled by another quantum system that can be in a superposition state. There are indeed extra advantages possible with this configuration that cannot be done with "regular" quantum computing.

This recent field is known as indefinite causal order. There have been some proofs that you cannot just set up a computer with a pure superposition of the gates acting in different orders; the only possibility is to have the order controlled by another quantum system. One can define things called causal inequalities (like Bell inequalities) to see whether a process has definite or indefinite causal order. And, these things can be used to find advantages in communication complexity, query complexity, etc. I'd start with the paper I linked at the outset and then perhaps follow other papers that cited it, as that one started the field (even though they cite Lucien Hardy with some similar ideas and even though it was on the arxiv for four years in which time other work was published).

One user I've seen on this site who's active in this area of research https://quantumcomputing.stackexchange.com/users/12541/mateus-ara%c3%bajo.

  • $\begingroup$ Oh interesting paper, I hadn't heard of this before! Thanks for the answer! $\endgroup$ Aug 17, 2023 at 7:23
  • $\begingroup$ @RajivKrishnakumar you are most welcome $\endgroup$ Aug 17, 2023 at 13:13

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